Question

The combustion of pentane, C5H12, occurs via the reaction C5H12(g)+8O2(g)→5CO2(g)+6H2O(g) with heat of formation values given by the following table: Substance ΔH∘f (kJ/mol) C5H12 (g) -119.9 CO2(g) −393.5 H2O(g) −241.8 Calculate the enthalpy for the combustion of pentane. Express your answer to four significant figures and include the appropriate units.

Answer 1

Answer:  -3298.4 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_5H_{12}(g)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{H_2O}\times \Delta H_{H_2O})+(n_{CO_2}\times \Delta H_{CO_2})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_5H_{12}}\times \Delta H_{C_5H_{12}})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$\Delta H=[(6\times -241.8)+(5\times -393.5)]-[(8\times 0)+(1\times -119.9)]$

$Delta H=-3298kJ/mol$

Therefore, the enthalpy change for this reaction is, -3298 kJ/mol